If tan^{-1}(x^2 + y^2) = alpha,then frac{dy}{ | vedclass

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Question

(A) Given that,$	an^{-1}(x^2 + y^2) = \alpha$. Taking $	an$ on both sides,we get $x^2 + y^2 = 	an \alpha$. Differentiating both sides with respect

Vedclass · Accepted Answer

$-\frac{x}{y}$

Vedclass · Answer

(A) Given that,$	an^{-1}(x^2 + y^2) = \alpha$. Taking $	an$ on both sides,we get $x^2 + y^2 = 	an \alpha$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(	an \alpha)$. Since $\alpha$ is a constant,$\frac{d}{dx}(	an \alpha) = 0$. So,$2x + 2y \frac{dy}{dx} = 0$. $2y \frac{dy}{dx} = -2x$. $\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$.

If $\tan^{-1}(x^2 + y^2) = \alpha$,then $\frac{dy}{dx}$ is equal to

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